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Derivada funciones vectoriales 2
 

Derivada funciones vectoriales 2

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    Derivada funciones vectoriales 2 Derivada funciones vectoriales 2 Document Transcript

    • Derivadas de Funciones Vectoriales 2Teorema 1. Si f es una funci´n vectorial continua sobre [a, b] y diferenciable sobre (a, b), oentonces existe Ci (a, b) tal que f (b) − f (a) = (b − a)(f1 (C1 ), . . . , fn (Cn )).Demostraci´n: Por hipotesis fi es continua y diferenciable, por lo tanto para cada fi existe o fi (b) − fi (a) Ci (a, b) tal que fi (Ci ) = . b−a por lo tanto fi (Ci )(b − a) = fi (b) − fi (a) por lo tanto f (b) − f (a) = (b − a)(f1 (C1 ), . . . , fn (Cn ))Teorema 2. Si ϕ es diferenciable sobre un intervalo I ⊂ R y f es diferenciable sobre unintervalo que contiene a ϕ(I) = {ϕ(t)|t I}, entonces f ◦ ϕ es diferenciable en I y (f ◦ ϕ) =f ◦ ϕ · ϕ sobre IDemostraci´n: Tenemos que: o (f ◦ ϕ) = [(f1 ◦ ϕ) , . . . , (fn ◦ ϕ) ] = [f1 ◦ ϕ · ϕ , . . . , fn ◦ ϕ · ϕ ] = [f1 ◦ ϕ, . . . , fn ◦ ϕ]ϕ =f ◦ϕ·ϕF´rmula de Taylor oSi f (t) = (f1 (t), f2 (t), ..., fn (t)) es una funci´n vectorial que es continua y diferenciable hasta on + 1 − veces entonces para cada i = 1, ..., n se tiene que h2 hn n fi (t + h) = fi (t) + h · fi (t) + + · fi (t) + ... + · f (t) + Rin 2! n! i ∴ f (t + h) = (f1 (t + h), f2 (t + h), ..., fn (t + h)) = h2 hn n h2 hn n(f1 (t)+h·f1 (t)++ ·f1 (t)+...+ ·f1 (t)+R1n , ..., fn (t)+h·fn (t)+ ·fn (t)+...+ ·fn (t)+R1n ) = 2! n! 2! n! 1
    • h2 hn n n (f1 (t), ..., fn (t)) + h · (f1 (t), ..., fn (t)) + (f1 (t), ..., fn (t)) + ... + (f1 (t), ..., fn (t)) 2! n! 2 n h h = f (t) + h · f (t) + f (t) + ... + f n (t) 2! n!que es la f´rmula de taylor para funciones vectoriales oDefinici´n 1. Si f = (f1 , . . . , fn ) es una funcion vectorial definida sobre [a, b], entonces o b b b f= f1 , . . . , fn . a a a bLa integral existe siempre que cada una de las integrales fi con i = 1, . . . , n existe. En a bparticular, si f es continua sobre [a, b] entonces f (t)dt existe. aTeorema 3. Si f = (f1 , . . . , fn ) es continua sobre un intervalo I y a I entonces: t d f a = f (t) ∀ t I dt Demostraci´n. La prueba se obtiene por la aplicaci´n del primer teorema fundamental o o del c´lculo a cada una de las funciones componentes a t d f t t a d = f1 , . . . , fn dt dt a a t t d d f1 , . . . , fn dt a dt a = (f1 (t), . . . , fn (t)) = f (t)Teorema 4. Si f = (f1 , . . . , fn ) tiene derivada continua sobre un intervalo I, entonces ∀ a, b I b f (t) = f (b) − f (a) a 2
    • Demostraci´n. o b b f = (f1 , . . . , fn ) a a t t = f1 , . . . , fn a a = (f1 (b) − f1 (a), . . . , fn (b) − fn (a)) = f (b) − f (a)Teorema 5. Si f (t) = (f1 (t), f2 (t), ..., fn (t)) es integrable en [a, b], para todo vector C =(c1 , c2 , ..., cn ) entonces el producto escalar C · F es integrable en [a, b] y b b C· f (t)dt = C · f (t)dt a aDemostraci´n. Tenemos que o b b b b C· f (t)dt = (c1 , c2 , ..., cn ) · f1 (t)dt, f2 (t)dt, ..., fn (t)dt = a a a a b b b c1 · f1 (t)dt + c2 · f2 (t)dt + ... + cn · fn (t)dt = a a a b b b b c1 · f1 (t)dt+, c2 · f2 (t)dt + ... + cn · fn (t)dt = C · f (t)dt a a a aTeorema 6. Si f y f son integrables en [a, b] entonces b b f (t)dt ≤ f (t) dt a a bDemostraci´n. Sea C = o a f (t)dt por lo que se tiene b b b b 2 C =C ·C =C · f (t)dt = C · f (t)dt ≤ |C · f (t)|dt ≤ C f (t) dt a a a a b b b b 2 ∴ C ≤ C f (t) dt ⇒ C ≤ f (t) dt ⇒ f (t)dt ≤ f (t) dt a a a a 3